abs_normal qp_box: Example and Test

qp_box.cpp

@(@\newcommand{\W}[1]{ \; #1 \; } \newcommand{\R}[1]{ {\rm #1} } \newcommand{\B}[1]{ {\bf #1} } \newcommand{\D}[2]{ \frac{\partial #1}{\partial #2} } \newcommand{\DD}[3]{ \frac{\partial^2 #1}{\partial #2 \partial #3} } \newcommand{\Dpow}[2]{ \frac{\partial^{#1}}{\partial {#2}^{#1}} } \newcommand{\dpow}[2]{ \frac{ {\rm d}^{#1}}{{\rm d}\, {#2}^{#1}} }@)@ abs_normal qp_box: Example and Test
Problem
Our original problem is @[@ \begin{array}{rl} \R{minimize} & x_0 - x_1 \; \R{w.r.t} \; x \in \B{R}^2 \\ \R{subject \; to} & -2 \leq x_0 \leq +2 \; \R{and} \; -2 \leq x_1 \leq +2 \end{array} @]@

Source


# include <limits>
# include <cppad/utility/vector.hpp>
# include "qp_box.hpp"

bool qp_box(void)
{     bool ok = true;
     typedef CppAD::vector<double> vector;
     //
     size_t n = 2;
     size_t m = 0;
     vector a(n), b(n), c(m), C(m), g(n), G(n*n), xin(n), xout(n);
     a[0] = -2.0;
     a[1] = -2.0;
     b[0] = +2.0;
     b[1] = +2.0;
     g[0] = +1.0;
     g[1] = -1.0;
     for(size_t i = 0; i < n * n; i++)
          G[i] = 0.0;
     //
     // (0, 0) is feasible.
     xin[0] = 0.0;
     xin[1] = 0.0;
     //
     size_t level   = 0;
     double epsilon = 99.0 * std::numeric_limits<double>::epsilon();
     size_t maxitr  = 20;
     //
     ok &= CppAD::qp_box(
          level, a, b, c, C, g, G, epsilon, maxitr, xin, xout
     );
     //
     // check optimal value for x
     ok &= std::fabs( xout[0] + 2.0 ) < epsilon;
     ok &= std::fabs( xout[1] - 2.0 ) < epsilon;
     //
     return ok;
}

Input File: example/abs_normal/qp_box.cpp